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This is my question:

Let P be a plane considered as a surface in 3-space. Show that its second fundamental form is zero. Conversely, show that if the second fundamental form of a surface is identically zero then the surface is a plane.

So far I have that the general equation of the plane is $$ax+by+cz+d=0.$$ Then $$M(u,v)=(u,v,\alpha u + \beta v + \gamma),$$ where $a,b,c,d,\alpha,\beta, \gamma$ are all constants and $c\not=0$. Then $M_u=(1,0,\alpha)$ and $M_v=(0,1,\beta)$. I know that $\vec{n} =\frac{M_u \times M_v}{|M_u \times M_v|}$ which gives $$ \vec{n}=\frac{(-\alpha,-\beta, 1)}{\sqrt{\alpha^2+\beta^2+1}}.$$ Hence, the unit norm is a constant. $$ M_{uu}=M_{uv}=M_{vv}=0$$ Hence $l=m=n=0$, so the second fundamental form is $$ ldu^2+mdudv+ndv^2=0.$$

Now I need to show the converse, but I don't know how. Please help :)

Lucy
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3 Answers3

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Proposition 4.2 If the second fundamental form of a surface vanishes, it is part of a plane.

Proof: If the second fundamental form vanishes,

$0 = r_u \cdot n_u = r_v \cdot n_u = r_u \cdot n_v = r_v \cdot n_v$

so that

$n_u = n_v = 0$

since $n_u$, $n_v$ are orthogonal to $n$ and hence linear combinations of $r_u$, $r_v$.

Thus $n$ is constant.

This means

$(r \cdot n)_u = r_u \cdot n = 0$,

$(r \cdot n)_v = r_v \cdot n = 0$

and so

$r \cdot n$ = const

which is the equation of a plane.

Emolga
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Dr. Math
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3

Hint: If the second fundamental form is always $0$, what does this tell you about $\vec n_u$ and $\vec n_v$?

Ted Shifrin
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0

do carmo page 147-148 or Luis Fernandez, Dep of Math, Univ of Bath, Differential geometry of curves and surfaces, page 52-53

Ali Ektiar
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